The following calculation note pertains to the E-906 Prop Tube Assembly, as developed by Ming Liu andWalt Sondheim.These calculations describe the ability of the Tube Assembly to handle service andhandling loads.

DESIGN:

Two rows of tubes are nested one on top of the other, but staggered. A bead of glue is placedlongitudinally at the tube to tube interface. This glue will provide additional structural support, but itwill not be taken credit for in the structural analysis for simplicity. Additionally,uncertainties regardingthe strength of the glue and its ability toadhere to the aluminum tube make it difficult to include in theanalysis.

ANALYSIS:

Refer to the Appendix in “Mechanical Engineering Design”, 4th

Edition, Shigley and Mitchell, formaximum moment, stress, and deflection formulas. Refer to Appendix C, “Mechanics of Materials”,Second Edition, Gere and Timoshenko for formula and explanation of the parallel axis theorem.

Calculation # 1

Analyze the one tube alone to determine if it can support

its own weight, simply supported at the ends.

Calculate weight of tube,

douter

= 2.00 in.

dinner= 1.93in.

ttube wall

= 0.035 in.

Weightsingle tube

= (cross sectional area)(length)(density)

Areatube cross section= π [(1.0 in.)2

–

(0.965 in.)2] = 0.2161 in.2

Densityaluminum

= 0.1 lb/in.3

Length = 144 in.

Weightsingle tube

=3.11 lb

Weightsingle tube/length=

0.0216 lb/in

Isingle tube

= π [(douter

)4

–

(dinner)4] / 64 = 0.1043 in.4

Maximum moment for a simply supported beam with a distributed load is

Mmax

= (w

l2)/8

= 56 in-lb

w

=3.11 lb/144 in. =0.0216 lb/in.

l = length = 144 in.

σmax

= M c/I = (56 in-lb)(1.00 in.)/0.1043 in.4

= 537 psi

c = tube radius = 1.00 in.

δmax

= (5wl4)/(384EI) = 0.12 in.

Ealuminum

= 10,000,000 psi

Tubes are aluminum 6061-T6with a yield stress of 35,000 psi, per ASTM B308. Thus, stress isacceptable. Deflection is small. Stressis also low enough so that local tube wall buckling should not bea concern.

Glued joints not taken credit for, and ends are clamped by the scalloped pieces. Thus, tubesindividually are acceptable.

Calculation #2

Calculate weight of tube array. Eachtube weighs 3.11 lb, and there are 144

tubes. Thus totaltubeweight is 450

lb.

Add 10% for glue. For analysis purposes, use:

Weighttube array

= (450 lb)1.1 = 495 lb

Weight of the structural members must be considered. Use a total

weight

(tubes and structural frame)

of 700 lb.

Calculation #3–

Lifting Features

Frame in vertical position

(tube axes perpendicular to floor)

A lifting plate and swivel hoist ring will be used at each corner. The plate will be 5/8 inch thick and madefrom 6061-T6. A

This stress is acceptable at it is less than the tensile allowable calculated in Calculation #9

below.

Frame in horizontal position

(tube axes parallel to floor)

The plate is secured to the frame with eight 0.25 inch diameter screws. Check shear strength of screws.

Ashear, 1/4” screw= 0.0269 in.2

τ = V/Ashear, eight 1/4” screw= (350 lb)/(8(0.0269 in.2

))= 1,626 psi

The shear stress is acceptable since it is less than the allowable calculated in Calculation #9.

Calculation #4–

Stresses in Frame Due To Lifting

The frame may be lifted from the horizontal position (frame parallel to floor) into the vertical position.

Frame in horizontal position

(tube axes parallel to floor)

The frame would be supported along two opposite edges of the frame. Total

frame weight is estimatedat 700 lb. This will be a distributed load between the two supported ends. Maximum bending momentwill occur between the supported ends.

Mmidspan

= wL2/8 or FL/8

Where:

L=length of span

w = distributed load

= 700 lb/152 in. = 4.6 lb/in.

F = total load

= wL

Mmidspan

= (700 lb)( 152 inches )/8 = 13,300 in-lb

Need to take credit for both top and bottom layers of 1020 structural members. This will mean thatshear must be transmitted effectively between top and bottom layers. Check bending stress forsituation where top and bottom 1020 layers participate. Use parallel axis theorem to calculate netmoment of inertia (Ix

No credit is taken for the tubes. Actual deflection will be less than this.

Maximum shear load occurs at each supported end, and is equal to 700 lb/2. Shear will be carried oneither side of the frame, so each side mustcarry 175 lb. Strap plates are 5.75 in. x 6.00 in. and

are 0.188thick. Shear stress is calculated as follows.

Ashear

= 6.00 in. x 0.188 in. = 1.13 in.2

τ = V/A = 175 lb/1.13 in.2

= 155 psi

Calculate stress in the screws which attach the strap plate to the 1020. Four 0.25 inch diameter screwsmust react both the actual shear force (175 lb/4 screws = 44 lb) as well as the couple developed by 175lb over a lever arm of 5 inches.

(175 lb)(5 in.) = 4(2.8 in.)Vcouple

(do not take credit for inner two screws as lever arm is short)

Vcouple

= 78 lb

Vlateral load

=175 lb/4 screws = 44 lb

Vtotal

= 78 lb + 44 lb = 122 lb

τ = Vtotal/A0.25 in. diam. screw

= 122 lb/0.0269 in.2

= 4,535 psi

Shear stress is acceptable since it is less than the allowable calculated in Calculation #9. The directionsof the two shear forces are 45 degrees apart, which will lower the stress. Additionally, the other threeplates inboard of the ends will help carry the shear as well. Shear stress calculated above isconservative.

Frame in vertical

position

(tube axes perpendicular to floor)

Stresses in the frame are minimal in the vertical position. Loads are well distributed throughout frame,no significant bending moments occur,and stresses are primarily membrane through the framemembers.

Loads are well distributed from the lifting lug locations.

Calculation #5

Calculate bending stress and shear stress for the situation where the frame is supported by a supportthat runs across the middle of the frame. In other words, frame

is set down on a 4” X 4” wood memberlying on the floor. The wood member runs across the frame mid-span perpendicular to the tube

axisdirection.

Attached 1020 information shows the moment of inertia about the weak axisof a 1020 memberto be0.0833in.4

The tubes are supported at the ends via the scalloped brackets.

The bottom 1020 pieces(one on each side) will initially be considered to carry the load. Use the case of a cantilevered beam

whose length

is 76 inches. Point load is 700 lb/4 or 175

lb.

Momentmax

= (175 lb)(76in) = 13,300 in-lb

σmax

= M c/I = (13,300 in-lb)(0.5in.)/0.0833 in.4

= 79,832

psi

Stress is too high, as yield stress is35 ksi

for 6105-T5 aluminum.

Need to take credit for both top and bottom layers. This will mean that shear must be transmittedeffectively between top and bottom layers. Check bending stress for situation where top and bottom1020 layers participate. Use parallel axis theorem to calculate net moment of inertia (Ix

= Ixc

+Ad12).

Itotal

=2(0.0833 in.4

)+2( 0.7914 in.2)(2 in.)2

= 6.5

in.4

σmax

= M c/I = (13,300 in-lb)(2.5in.)/6.5

in.4

= 5,115

psi

Bending stress is acceptable. Reaction load at ends is 175 lb, so shear force of 175

lb must be carriedwithin the frame at the ends. Strap plates are 5.75 in. x 6.00 in. and are 0.188 thick. Shear area is

Ashear

= 6.00 in. x 0.188 in. = 1.13 in.2

τ= V/A = 175

lb/1.13 in.2

=

155

psi

Calculate stress in the screws which attach the strap plate to the 1020. Four

0.25 inch diameter screwsmust react both the actual shear force as well as the couple developed by 175

lb over a lever arm of5inches.

(175lb)(5 in.) =4(2.8in.)Vcouple

(do not take credit for inner two screws as lever arm is short)

Vcouple

= 78

lb

Vlateral load

=175 lb/4 screws = 44

lb

Vtotal

=78 lb + 44 lb = 122

lb

τ = Vtotal/A0.25 in. diam. screw

= 122

lb/0.0269

in.2

=

4,535

psi

Stress is acceptable

at is less than the allowable calculated in Calculation #9

below. The directions of thetwo shear forces are 45

degrees apart, which will

lower the stress.Additionally,

the other three platesinboard of the ends will help carry the shear as well.Shear stress calculated above is conservative.

Calculation #6

This calculation is similar to Calculation #5

above, but the wood beam placedon the floor runscoincident with the tubes. It supports the frame in the middle, but 90 degrees to the wood

floorsupport in Calculation #5. The frame is square and there are more support members running across theends so stresses will be lower.

Momentmaximum

= 13,300 in-lb (from above, as frame is square)

Frame has a lower layer and an upper layer. Strap plates will connect the upper and lower layers, sothat the upper and lower layers can be taken credit forin bending, as in Calculation #5.

There is a 1020and 1030 piece on the bottom, and a 1020 and 1030 piece on top. The effective I for these bottom andtop layers

is more than for Calculation #5. The maximum moment and c are the same. Thus bendingstress will be lower. Shear will betransmitted as in Calculation #5, so shear stress is acceptable.

Scalloped pieces were not taken credit for.

Also this case involves more of a distributed load alongbeam, versus a point load at the end

for Calculation #5.

Calculation #7

Check bendingstress for bending about an axis that runs

corner to corner. Frame laid down on flooronto wood beam from Calculation #5 and #6, wherein beam runs from frame corner to frame corner.Use parallel axis theorem again, and use the plate thickness at the corners.

Calculate maximum moment. The areaon either side of the wood beamis triangular, with the center ofmass 1/3 of the way out from the diagonal bending axis, or 108.2 in./3 or 36 inches.

Momentmaximum

= (350 lb)(36 in.) = 12,600 in.-lb

Use parallel axis theorem

to calculate net moment of inertia (Ix

= Ixc

+Ad12)

for the brackets at the framecorners.

Itotal

for brackets at corners

=~0 in.4

+ 8(0.188 in.)( 4.0 in.)(2.5 in.)2

= 37.6

in.4

σmax

= M c/I = (12,600 in-lb)(2.5in.)/37.6

in.4

= 838

psi

Check shear stress in screws that attach the eight plates.

Momentmaximum

= (5in.)(4 plates)Fplate

(summing moments about lower frame surface at brackets)

Fplate

=630 lb.

Each plate has at least 12 screws. Each sideof the platehas six screws. Force per screw is then 630 lb/6or 105

lb. Screws are ¼”-20, with a shear area of 0.0269 in.2

The shear stress in the screw is calculatedas follows.

τ = Vtotal/A0.25 in. diam. screw

= 105

lb/0.0269 in.2

=

3,904

psi

Shear stress is acceptable as it is less than the allowable calculated in Calculation #9

below.

Calculation #8

The following calculation will consider frame skewing (i.e. frame tendency to go from a square to aparallelogram shape). This deformation mode is what thefour diagonal members are used to prevent.

This scenario would cover an event where the frame is lowered

onto one corner while providing lateralrestraint.

The following will calculate the stress in the diagonal, the attachment plate and the screws.

As the frame tries to skew,

two of the four top diagonals will be placed in tension, and two incompression. The same holds true for the bottom layer. Only the diagonals placed in tension will betaken credit for since

the compressed two could bow. Refer to Sketch 1below.

The total frame weight is 700 lb. Frame corner connections will be assume to be pinned, with all of thesupport provided by the diagonals. Half of the load carried by the bottom layer of four diagonals , andhalf carried by the top layer. Each top layertensile diagonal then carries 175

lb

(700 lb /2 layers is 350 lbper layer, then divided by two for two tensile diagonals is 175 lb

The diagonal is made from 1010 with a cross sectional area of 0.4379 in.2

σbrace

= F/A =250

lb/0.4379 in.2

= 571

psi

The bracket attaching the diagonal is 0.188 in. thick with a min. width of 2 in.

Abracket, cross section

= (0.188 in.)(2 in.) = 0.376 in.2

σbracket

= F/A = 250

lb/0.376 in.2

= 665

psi

Bracket is attached with two screws on either side. Shear stress in screws is calculated as follows.

Ashear, 0.25 in. diam.=0.0269 in.2

τ = Vtotal/2A0.25 in. diam. screw

= 250

lb/[2(0.0269 in.2

)] = 4,647

psi

The same results hold for the other top layer diagonal,

as each is assumed to carry 175

lb.

The calculated stress in the screw is less than the allowable calculated in Calculation #9, and is thereforeacceptable.

Calculation #9

The strength of the fastened connection between joining plates and the 1020 or 1030 members will beevaluated herein. The fasteners which attach the joining plates to the 1020 or 1030 members are ¼”-20screws. According to the manufacturer,

these screws are made from SAE Grade 8 material.

Thesescrews go through the joining plates and are threaded into an “economy nut” within the 1020 or 1030channel. It is difficult to analyze the shear strength(load across fastener) or tensile strength (load alongfastener axis) of this connection

with respect to thenut pulling out of the extruded part. Optimisticallyone would consider the shear strengthor tensile strength of the screw (Grade 8 material)

and use alarge allowable shear force

or tensile force. The 80/20 manufacturer recommends an allowableshearload of 175 lb per the attached page 155 of their catalog. The shear area of a ¼”-20 screw is 0.0269 in.2.The allowable shear stress cold be calculated using the vendor load as,

τallowable

= Vvendor allowable

/ Ashear

= 175 lb/0.0269 in.2

= 6,505 psi

This pertains to shear loading of the screw.This stress will be treated as the maximum allowable shearstress in the ¼”-20 screws for the frame analysis.This stress will also be treated as the maximumallowable tensile stress for the screws

for the frame analysis.

For reference,

thetensileyield strength of SAE Grade 8 material is 130,000 psi,

per SAE J429. The shearyield strength is taken as 60% of this value, or 78,000 psi. The shear and tensile allowable stress of6,505 psi usedfor the frame analysis ismuch lower(order of magnitude)thanthe78,000 psi (shear) or130,000 psi (tensile)